原題

給出一棵二叉樹，返回其節(jié)點(diǎn)值的前序遍歷。

給出一棵二叉樹 {1,#,2,3}

   1
    \
     2
    /
   3

返回 [1,2,3]

解題思路

• helper函數(shù)中if root:即表明root如果不是None，就進(jìn)行父/左/右的添加
• 方法一，最簡(jiǎn)單，recursion (no return value)
• 方法二，divide & conquer (has return value)
• 方法三，非遞歸，因?yàn)檫f歸調(diào)用了系統(tǒng)中的?？臻g，所以非遞歸的實(shí)現(xiàn)要用到stack這種數(shù)據(jù)結(jié)構(gòu)

完整代碼

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# 方法一
class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        ret = []
        self.RecPreOrderTraversal(root, ret)
        return ret
        
    def RecPreOrderTraversal(self, root, result):
        if root != None: # skip None or Leaf
            result.append(root.val)
            self.RecPreOrderTraversal(root.left, result)
            self.RecPreOrderTraversal(root.right, result)

# 方法二
class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        if root is None:
            return res
        
        left = self.preorderTraversal(root.left)
        right = self.preorderTraversal(root.right)
        
        res.append(root.val)
        res += left
        res += right
        return res 

# 方法三
class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack = [root]
        preorder = []
        while stack:
            node = stack.pop()
            preorder.append(node.val)
            if node.right:
                stack.append(node.right)
            if node.left:
                stack.append(node.left)
        return preorder